For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation $\begin{array} { c }
\sigma ^ { \wedge } \
p
\end{array}$
A) 0.96
B) 0.19
C) 0.040
D) 0.81

Question

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation $\begin{array} { c } 
\sigma ^ { \wedge } \
p
\end{array}$
A) 0.96
B) 0.19
C) 0.040
D) 0.81

Quizplus · Accepted Answer

Since we are dealing with a binomial distribution (either a diamond qualifies as gemstone grade or it doesn't), the formula for the standard deviation is:
$$
\sigma = \sqrt{np(1-p)}
$$
where n is the sample size and p is the probability of success (qualifying as gemstone grade) in a single trial.

In this case, n = 97 and p = 0.19 (since 81% of diamonds fail to qualify, then 19% qualify), so plugging in these values:
$$
\sigma = \sqrt{(97)(0.19)(0.81)} \approx 0.04
$$

Therefore, the best choice is C.

For a Particular Diamond Mine, 81% of the Diamonds Fail σ∧p\begin{array} { c } \sigma ^ { \wedge } \\p\end{array}σ∧p​

For a Particular Diamond Mine, 81% of the Diamonds Fail $\begin{array} { c } \sigma ^ { \wedge } \\p\end{array}$